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Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

The length of the diameter of the circle which touches the $$x$$-axis at the point $$(1, 0)$$ and passes through the point $$(2, 3)$$ is:

A

$${{10} \over 3}$$

B

$${{3} \over 5}$$

C

$${{6} \over 5}$$

D

$${{5} \over 3}$$

Let center of the circle be $$\left( {1,h} \right)$$

$$\left[ {\,\,} \right.$$ as circle touches $$x$$-axis at $$\left. {\left( {1,0} \right)\,\,} \right]$$

Let the circle passes through the point $$B(2,3)$$

$$\therefore$$ $$CA=CB$$ (radius)

$$ \Rightarrow C{A^2} = C{B^2}$$

$$ \Rightarrow {\left( {1 - 1} \right)^2} + \left( {h - 0} \right){}^2 = {\left( {1 - 2} \right)^2} + {\left( {h - 3} \right)^2}$$

$$ \Rightarrow {h^2} = 1 + {h^2} + 9 - 6h$$

$$ \Rightarrow h = {{10} \over 6} = {5 \over 3}$$

Thus, diameter is $$2h = {{10} \over 3}.$$

$$\left[ {\,\,} \right.$$ as circle touches $$x$$-axis at $$\left. {\left( {1,0} \right)\,\,} \right]$$

Let the circle passes through the point $$B(2,3)$$

$$\therefore$$ $$CA=CB$$ (radius)

$$ \Rightarrow C{A^2} = C{B^2}$$

$$ \Rightarrow {\left( {1 - 1} \right)^2} + \left( {h - 0} \right){}^2 = {\left( {1 - 2} \right)^2} + {\left( {h - 3} \right)^2}$$

$$ \Rightarrow {h^2} = 1 + {h^2} + 9 - 6h$$

$$ \Rightarrow h = {{10} \over 6} = {5 \over 3}$$

Thus, diameter is $$2h = {{10} \over 3}.$$

2

MCQ (Single Correct Answer)

The two circles x^{2} + y^{2} = ax, and x^{2} + y^{2} = c^{2} (c > 0) touch each other if

A

| a | = c

B

a = 2c

C

| a | = 2c

D

2 | a | = c

As center of one circle is $$\left( {0,0} \right)$$ and other circle passes through $$(0,0),$$ therefore

Also $${C_1}\left( {{a \over 2},0} \right){C_2}\left( {0,0} \right)$$

$${r_1} = {a \over 2}{r_2} = C$$

$${C_1}{C_2} = {r_1} - {r_2} = {a \over 2}$$

$$ \Rightarrow C - {a \over 2} = {a \over 2}$$

$$ \Rightarrow C = a$$

If the two circles touch each other, then they must touch each other internally.

Also $${C_1}\left( {{a \over 2},0} \right){C_2}\left( {0,0} \right)$$

$${r_1} = {a \over 2}{r_2} = C$$

$${C_1}{C_2} = {r_1} - {r_2} = {a \over 2}$$

$$ \Rightarrow C - {a \over 2} = {a \over 2}$$

$$ \Rightarrow C = a$$

If the two circles touch each other, then they must touch each other internally.

3

MCQ (Single Correct Answer)

The circle $${x^2} + {y^2} = 4x + 8y + 5$$ intersects the line $$3x - 4y - m$$ at two distinct points if

A

$$ - 35 < m < 15$$

B

$$ 15 < m < 65$$

C

$$ 35 < m < 85$$

D

$$ - 85 < m < -35$$

Circle $${x^2} + {y^2} - 4x - 8y - 5 = 0$$

Center $$=(2,4),$$ Radius $$ = \sqrt {4 + 16 + 5} = 5$$

If circle is intersecting line $$3x-4y=m,$$ at two distinct points.

$$ \Rightarrow $$ length of perpendicular from center to the line $$ < $$ radius

$$ \Rightarrow {{\left| {6 - 16 - m} \right|} \over 5} < 5 \Rightarrow \left| {10 + m} \right| < 25$$

$$ \Rightarrow - 25 < m + 10 < 25 \Rightarrow - 35 < m < 15$$

Center $$=(2,4),$$ Radius $$ = \sqrt {4 + 16 + 5} = 5$$

If circle is intersecting line $$3x-4y=m,$$ at two distinct points.

$$ \Rightarrow $$ length of perpendicular from center to the line $$ < $$ radius

$$ \Rightarrow {{\left| {6 - 16 - m} \right|} \over 5} < 5 \Rightarrow \left| {10 + m} \right| < 25$$

$$ \Rightarrow - 25 < m + 10 < 25 \Rightarrow - 35 < m < 15$$

4

MCQ (Single Correct Answer)

If $$P$$ and $$Q$$ are the points of intersection of the circles

$${x^2} + {y^2} + 3x + 7y + 2p - 5 = 0$$ and $${x^2} + {y^2} + 2x + 2y - {p^2} = 0$$ then there is a circle passing through $$P,Q $$ and $$(1, 1)$$ for:

$${x^2} + {y^2} + 3x + 7y + 2p - 5 = 0$$ and $${x^2} + {y^2} + 2x + 2y - {p^2} = 0$$ then there is a circle passing through $$P,Q $$ and $$(1, 1)$$ for:

A

all except one value of $$p$$

B

all except two values of $$p$$

C

exactly one value of $$p$$

D

all values of $$p$$

The given circles are

$${S_1} \equiv {x^2} + {y^2} + 3x + 7y + 2p - 5 = 0\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

$${S_2} \equiv {x^2} + {y^2} + 2x + 2y - {p^2} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$

$$\therefore$$ Equation of common chord $$PQ$$ is $${S_1} - {S_2} = 0$$

$$ \Rightarrow L \equiv x + 5y + {p^2} + 2p - 5 = 0$$

$$ \Rightarrow $$ Equation of circle passing through $$P$$ and $$Q$$ is

$${S_1} + \lambda \,\,L = 0$$

$$ \Rightarrow \left( {{x^2} + {y^2} + 3x + 7y + 2p - 5} \right) + \lambda $$

$$\left( {x + 5y + {p^2} + 2p - 5} \right) = 0$$

As it passes through $$\left( {1,1} \right),$$ therefore

$$ \Rightarrow \left( {7 + 2p} \right) + \lambda \left( {2p + {p^2} + 1} \right) = 0$$

$$ \Rightarrow \lambda = - {{2p + 7} \over {\left( {p + 1} \right)}},$$

which does not exist for $$p=-1$$

$${S_1} \equiv {x^2} + {y^2} + 3x + 7y + 2p - 5 = 0\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

$${S_2} \equiv {x^2} + {y^2} + 2x + 2y - {p^2} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$

$$\therefore$$ Equation of common chord $$PQ$$ is $${S_1} - {S_2} = 0$$

$$ \Rightarrow L \equiv x + 5y + {p^2} + 2p - 5 = 0$$

$$ \Rightarrow $$ Equation of circle passing through $$P$$ and $$Q$$ is

$${S_1} + \lambda \,\,L = 0$$

$$ \Rightarrow \left( {{x^2} + {y^2} + 3x + 7y + 2p - 5} \right) + \lambda $$

$$\left( {x + 5y + {p^2} + 2p - 5} \right) = 0$$

As it passes through $$\left( {1,1} \right),$$ therefore

$$ \Rightarrow \left( {7 + 2p} \right) + \lambda \left( {2p + {p^2} + 1} \right) = 0$$

$$ \Rightarrow \lambda = - {{2p + 7} \over {\left( {p + 1} \right)}},$$

which does not exist for $$p=-1$$

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Complex Numbers

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